First slide

Planes in 3D

Question

The distance of the point (1,0,2) from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ is

Moderate

A

$2 \sqrt{14}$
B

8
C

$3 \sqrt{21}$
D

13

Solution

$\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12} = t$
$P (3 t + 2, 4 t - 1, 12 t + 2)$ lies on $x - y + z = 16 \Rightarrow t = 1$
$P (5, 3, 14) A (1, 0, 2)$

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